Order of transformation of functions?

In summary, when performing transformations on a function, it is important to follow the order of operations. First, compressions or expansions should be done, followed by reflections across an axis, and finally translations. Doing the transformations in any other order may result in incorrect values for the new function.
  • #1
Mr Davis 97
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I am confused about the order in which we apply transformations to a input of a parent function to get the corresponding input of the new function. Say for example, we have the function ##y = \sin(-2x + 1)=\sin(-2(x-\frac{1}{2}))##. Intuitively, it would seem as though we would transform a point from the parent function ##y=\sin(x)## by doing the following:
(4, - 0.7568)
(4.5, - 0.7568)
(2.25, - 0.7568)
(-2.25, - 0.7568)

However, when -2.25 radians is inputted into the new function, we get - 0.7055!

What am I doing wrong? I'm transforming the point as if it were an input to the function, but this is not producing the correct value. This begs the question: what is the order in which we apply transformations in order to get the correct value in the new function?
 
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  • #2
Mr Davis 97 said:
I am confused about the order in which we apply transformations to a input of a parent function to get the corresponding input of the new function. Say for example, we have the function ##y = \sin(-2x + 1)=\sin(-2(x-\frac{1}{2}))##. Intuitively, it would seem as though we would transform a point from the parent function ##y=\sin(x)## by doing the following:
(4, - 0.7568)
(4.5, - 0.7568)
(2.25, - 0.7568)
(-2.25, - 0.7568)

However, when -2.25 radians is inputted into the new function, we get - 0.7055!
It's hard to tell what you're doing with the points above.
When you're analyzing the various transformations made to an untransformed function, look at things in this order:
1. Compressions or expansions
2. Reflections across an axis (or across the origin if there are two such reflections)
3. Translations

I believe that the compressions/expansions and reflections can be performed in either order, but the translations have to be last.

For your function, the base function is y = sin(x).
1. y = sin(2x) compresses the base function toward the x-axis.
2. y = sin(-2x) reflects the graph just above across the x-axis.
3. y = sin(-2(x - 1/2)) translates the graph in #2 1/2 unit to the right.

One of the points on the base graph is ##(\pi/2, 1)##.
In #1, this point goes to ##(\pi/4, 1)##.
In #2, the point above goes to ##(-\pi/4, 1)##.
Finally, the point above goes to ##(-\pi/4 + 1/2, 1)##.

It's easy to verify that if ##-\pi/4 + 1/2## is the input to y = sin(-2(x - 1/2)), the output is 1.

Mr Davis 97 said:
What am I doing wrong? I'm transforming the point as if it were an input to the function, but this is not producing the correct value. This begs the question: what is the order in which we apply transformations in order to get the correct value in the new function?
 
  • #3
If I understand you correctly, you are starting with sin(4) and trying to find the x such that -2x+1 = 4?
That doesn't seem like a challenging problem.
Transformations start from the outside in.
If you have ##x' = g(h(x))##, then ##g^{-1}(x') = h(x)## and ##h^{-1} (g^{-1}(x')) = x##.
So, in your example ##\sin(x') = \sin( -2( x-1/2)) = \sin(g(h(x)))##
with:
##g(x) = -2x##
##h(x) = x-1/2##
having inverses:
##g^{-1}(x) = -x/2##
##h^{-1}(x) = x+1/2##
So looking for an x, given that x' = 4, ##x=h^{-1} (g^{-1}(x')) = -4/2 + 1/2 = 1.5 ##.
 
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  • #4
Mark44 said:
It's hard to tell what you're doing with the points above.
When you're analyzing the various transformations made to an untransformed function, look at things in this order:
1. Compressions or expansions
2. Reflections across an axis (or across the origin if there are two such reflections)
3. Translations

I believe that the compressions/expansions and reflections can be performed in either order, but the translations have to be last.

For your function, the base function is y = sin(x).
1. y = sin(2x) compresses the base function toward the x-axis.
2. y = sin(-2x) reflects the graph just above across the x-axis.
3. y = sin(-2(x - 1/2)) translates the graph in #2 1/2 unit to the right.

One of the points on the base graph is ##(\pi/2, 1)##.
In #1, this point goes to ##(\pi/4, 1)##.
In #2, the point above goes to ##(-\pi/4, 1)##.
Finally, the point above goes to ##(-\pi/4 + 1/2, 1)##.

It's easy to verify that if ##-\pi/4 + 1/2## is the input to y = sin(-2(x - 1/2)), the output is 1.

Okay, so now I understand that the order in which you perform the transformations to a test point. However, why must this be the order? For example, given the function ##
y = \sin(-2x + 1)=\sin(-2(x-\frac{1}{2}))## I would be inclined to take the point, ##(\pi/2, 1)##, from the parent function, and start from the inside out such that I do
$$(\pi/2 + 1/2, 1)$$
$$(\pi/4 + 1/4, 1)$$
$$(-\pi/4 - 1/4, 1)$$

Of course this leads to the incorrect answer, but why does it? Essentially, I want to know why so that I don't make the mistake of doing it.
 
  • #5
Mr Davis 97 said:
Okay, so now I understand that the order in which you perform the transformations to a test point. However, why must this be the order?
The short answer is, because if you do the translations before the compressions/expansions, you get the wrong answer.

Let's consider a slightly simpler example, y = f(x) = sin(2(x - 1/2))
The point (0, 0) is on the graph of y = sin(x).
If we look at the translation first (which is not the correct thing to do), we're looking at y = sin(x - 1/2), a translation to the right by 1/2 unit of the graph of y = sin(x). So it will contain the point (1/2, 0).

Next, we look at the compression. The factor of 2 should compress the graph of y = sin(x - 1/2) by a factor of 2 toward the y-axis. That would give us the point (1/4, 0).

A quick check shows that this is wrong. According to our calculations along the way, it should be that f(1/4) = 0.

f(1/4) = sin(2(1/4 - 1/2)) = sin(2(-1/4)) = sin(-1/2) ≠ 0.

This shows that we did something wrong.

OTOH, if we look at the compression first, we're looking at y = sin(2x), which should still have (0, 0) on its graph.
Now look at the translation, or y = sin(2(x - 1/2)). This should go through (1/2, 0).

A quick check shows that f(1/2) = sin(2(1/2 - 1/2)) = sin(0) = 0

As a sanity check, let's check another point on the graph of y = sin(x); namely, ##(\pi/2, 1)##.
1. Compression: ##(\pi/4, 1)## should be on the graph of y = sin(2x).
2. Translation: ##(\pi/4 + 1/2, 1)## should be on the graph of y = sin(2(x - 1/2)).
The check: ##f(\pi/4 + 1/2, 1) = sin(2(\pi/4 + 1/2 - 1/2)) = sin(2(\pi/4)) = sin(\pi/2) = 1##
Success!

I'll leave it for you to verify that starting with y = sin(x) and performing the transformations in the opposite order gives an incorrect result. As to why this is so, I've always been content to know what is the right way to look at these transformations, and haven't given much thought as to why one way works and another way doesn't. Possibly there is someone else here who has given it more thought.
Mr Davis 97 said:
For example, given the function ##
y = \sin(-2x + 1)=\sin(-2(x-\frac{1}{2}))## I would be inclined to take the point, ##(\pi/2, 1)##, from the parent function, and start from the inside out such that I do
$$(\pi/2 + 1/2, 1)$$
$$(\pi/4 + 1/4, 1)$$
$$(-\pi/4 - 1/4, 1)$$

Of course this leads to the incorrect answer, but why does it? Essentially, I want to know why so that I don't make the mistake of doing it.
 
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  • #6
RUber said:
If you have ##x' = g(h(x))##, then ##g^{-1}(x') = h(x)## and ##h^{-1} (g^{-1}(x')) = x##.
So, in your example ##\sin(x') = \sin( -2( x-1/2)) = \sin(g(h(x)))##
with:
##g(x) = -2x##
##h(x) = x-1/2##
having inverses:
##g^{-1}(x) = -x/2##
##h^{-1}(x) = x+1/2##
So looking for an x, given that x' = 4, ##x=h^{-1} (g^{-1}(x')) = -4/2 + 1/2 = 1.5 ##.
The explanation of why is given in my post above. Functions often do not commute. g(h(x)) is not equivalent to h(g(x)). And thus taking inverse a of these functions also is not give a commutative option.
If x' = g(h(x)) and you take the inverse of h first, you get
##h^{-1}(x') = h^{-1}(g(h(x))) ##
This is not helping you find x.
For simple problems, it is easy enough to work out the answer algebraically.
Given some x', f(x') = f( -2x+1), then x'= -2x+1. Solve for x.
 
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1. How do you determine the order of transformation for a function?

The order of transformation for a function is determined by the order in which the transformations are applied to the original function. For example, if the original function is f(x) and it is first translated horizontally by h units and then vertically by k units, the order of transformation would be f(x + h) + k. This is because the horizontal transformation is applied first, followed by the vertical transformation.

2. What is the difference between horizontal and vertical transformations?

Horizontal transformations involve shifting the graph of a function left or right along the x-axis, while vertical transformations involve shifting the graph up or down along the y-axis. Horizontal transformations are represented by the variable h and vertical transformations are represented by the variable k.

3. Can the order of transformation be changed?

Yes, the order of transformation can be changed without affecting the final result. This is because the order in which the transformations are applied does not change the final position of the graph, only the method in which it gets there.

4. How do you graph a function with multiple transformations?

To graph a function with multiple transformations, you can use the rule of translation. This means that you first graph the original function and then apply the transformations in the order specified. For example, if the function is f(x) and the transformations are f(x + h) + k, you would first graph f(x) and then shift it horizontally by h units and vertically by k units.

5. How can you use the order of transformation to simplify a function?

The order of transformation can be used to simplify a function by combining multiple transformations into one. This can be done by finding the final transformation rule using algebraic methods, such as substitution or elimination. Once the final transformation rule is found, it can be applied to the original function to simplify it.

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